Task 1. Paper Shell (box.pas/box.c/box.cpp)"Title description"MCX is a child with a mild cleanliness. One day, when he was addicted to the math paper difficult to extricate themselves, vaguely think of their own learning probability of the time to prepare a pile of rubber also cluttered in the study room. This is clearly beyond his tolerance. So he decided to make a paper box with a volume of V (V=ABC) so that it could be placed neatly. For simplicity, the length, width, and height of the paper box are positive integers. Of course, MCX is a thrifty kid, so he wants to know what is the minimum surface area (S=2AB+2AC+2BC) of this paper box? "topic input"a single row, which is a positive integer V, represents the volume of the paper box. "topic output"a single row, which is a positive integer s, represents the minimum surface area of the paper box. "Sample input 1" -"Sample Output 1" -"Sample input 2"146"Sample Output 2"442"Sample Interpretation"The only way to make a box with a volume of 17 is to have a length of 1 width 1 high 17 (here we can think long, wide, high is equivalent), so its minimum surface area is 2*1*17+2*1*17+2*1*1 ="Data range"30% data Meets v<=1000100% data Meets V<=10^9Analysis:First of all! This is one! Math problem!! The math problem is going to have a math problem--it's much better than the other number-theoretic flirtatious problem. There is really no advanced solution to this problem. But to think about it is actually an enumeration. Enum A, after enumeration B, C comes out naturally. and just the method, blind basic, small data should be able to get points. the first optimization is to enumerate the A<=pow (V, 1.0/3) only when enumerating a, and enumerate only the preceding. and b is enumerated b<=sqrt (v/a);C can be calculated directly. Of course, it is not an integer judgment. The second optimization is when enumerating a. A pruning judgment on whether to enumerate B. We know that because at the time of enumeration A. V and a are fixed values. The surface area formula is the minimum value of the surface area of the v/a+a* (AB+BC+AC), which is changed into a (B+C) (b+c). We also know V/A==BC so BC is also fixed value so when b==c is the theoretical minimum value. So if we find a theoretical minimum, it's bigger than the answer we've enumerated. Then there's no need to enumerate B. Why B*c when B==c (b+c) is the smallest when it is a certain time? Here is a special thanks to the explanation of the Big uncle:
Post the Code:
#include <cstdio> #include <algorithm> #include <cmath>using namespace Std;int main () { freopen (" Box.in "," R ", stdin); Freopen ("Box.out", "w", stdout); int V; Long long int ans=1000000000000000ll,bmj; scanf ("%d", &v); for (int a=v;a>=1;--a) { if ((v/a) *a!=v) continue; int kk=sqrt (v/a); bmj= (Long long int) (V/A+A*2*KK); if (bmj<=ans| | A==V) {for (int b=a;b<=a+kk;++b) { int c=v/a/b; if (c*a*b!=v) continue; Ans=min (ans, (long long int) a*b+ (long long int) b*c+ (long long int) a*c); }}} printf ("%i64d", 2*ans); Fclose (stdin); Fclose (stdout); return 0;}
NOIP Simulation Paper Shell
